Question

An emf source with E = 120 V, a resistor with R = 80.0 Ω, and a capacitor with C = 4.00 µF are connected in series. As the capacitor charges, when the current in the resistor is 0.900 A, what is the magnitude of the charge on each plate of the capacitor?

Short answer

When an emf source with E=120 V and a resistor with $80\mathrm{\Omega }$and a capacitor of $4.00\mathrm{\mu F}$are connected in series and the capacitor gets charges when the current in the resistor is 0.900 A then the magnitude of the charge in each plate capacitor is$192\mathrm{\mu C}$

Step-by-step solution

EMF of the battery

Electromotive force (Emf) is the total amount of electrical potential energy a battery can supply before it is connected to an external circuit

Potential difference across the circuit

We know that the charges Q on the plates of the capacitor depend on the voltage${\mathrm{V}}_{\mathrm{C}}$ across both the plates and it is given by:

$\mathrm{Q}={\mathrm{CV}}_{\mathrm{C}}$

Where C is the capacitance. To find the value of Q we need to determine ${\mathrm{V}}_{\mathrm{C}}$. To determine the${\mathrm{V}}_{\mathrm{C}}$we will use the loop rule.

$$\begin{gathered} \mathrm{\epsilon }-\mathrm{IR}-{\mathrm{V}}_{\mathrm{C}}=0 \\ {\mathrm{V}}_{\mathrm{C}}=\mathrm{\epsilon }-\mathrm{IR} \end{gathered}$$

Calculating the Voltage

Now in the above equation we put the values and get:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{C}}=\mathrm{\epsilon }-\mathrm{IR} \\ =(120\mathrm{V})-(0.900\mathrm{A})(80\mathrm{\Omega }) \\ =48\mathrm{V} \end{gathered}$$

Now, putting the values of ${\mathrm{V}}_{\mathrm{C}}$and C into the equation we get:

$$\begin{gathered} \mathrm{Q}={\mathrm{CV}}_{\mathrm{C}} \\ =(4\mathrm{\mu F})(48\mathrm{V}) \\ =192\mathrm{\mu C} \end{gathered}$$

Therefore, the magnitude of the charge is$192\mathrm{\mu C}$